The chord AB of the circle with center O is perpendicular to its radius OC

The chord AB of the circle with center O is perpendicular to its radius OC and divides it into segments OK = 7.5 and CK = 1. Find the length of chord AB.

Let’s draw the segments OA and OB from the center of the circle to the edges of the chord AB, the segments OA and OB are the radii of the circle.

The segment OK is also the radius of the circle, OC = R = OK + CK = 7.5 + 1 = 8.5 cm.

Then in the right-angled triangle AOK, by the Pythagorean theorem, we define the leg AK.

AK ^ 2 = OA ^ 2 – OK ^ 2 = 8.5 ^ 2 – 7.5 ^ 2 = 72.25 – 56.25 = 16.

AK = 4 cm.

Since the triangle AOB is isosceles, then the height OK is also the median of the triangle, then AK = BK = AB / 2.

AB = 2 * AK = 2 * 4 = 8 cm.

Answer: The length of the chord AB is 8 cm.